Sum of two opposite inequalities

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Sum of two opposite inequalities

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#1 h Jun 9, 2016, 1:40 AM • 2 Y

Y by Adventure10, Mango247

P5. Are well known $\frac {8abc}{(a+b)(b+c)(c+a)}\le\ 1\le \frac {a^2+b^2+c^2}{ab+bc+ca}$ for any $\{a,b,c\}\subset\mathrm R_+^*\ .$

Prove that $\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)}\ \ge\ 2\ .$

This post has been edited 8 times. Last edited by Virgil Nicula, Jun 9, 2016, 9:30 PM

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#2 h Jun 9, 2016, 2:18 AM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

P5. Are well known $\frac {8abc}{(a+b)(b+c)(c+a)}\le\ 1\le \frac {a^2+b^2+c^2}{ab+bc+ca}$ for any $\{a,b,c\}\subset\mathrm R_+^*\ .$ Prove that $\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)}\ \ge\ 2\ .$

Equivalent to $\dfrac{a^2+b^2+c^2}{ab+bc+ca}-1\geq 1-\dfrac{8abc}{(a+b)(b+c)(c+a)}$
$\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{2(ab+bc+ca)}\geq \dfrac{a(b-c)^2+c(a-b)^2+b(a-c)^2}{(a+b)(b+c)(c+a)}$
$\sum (b-c)^2\left[\dfrac{1}{2(ab+bc+ca)}-\dfrac{a}{(a+b)(b+c)(c+a)}\right]\geq 0$
Setting $S_a=\dfrac{1}{2(ab+bc+ca)}-\dfrac{a}{(a+b)(b+c)(c+a)}$ , similar to $S_b,S_c$
Without loss of genenerality, assume $a\geq b\geq c$ , hence $S_b=\dfrac{ab(a-b)+c(a^2-b^2)+c^2(a+b)}{2(ab+bc+ca)(a+b)(b+c)(c+a)}\geq 0$
And $S_c=\dfrac{(a+b)(ab-c^2)+c(a^2+b^2)}{2(ab+bc+ca)(a+b)(b+c)(c+a)}\geq 0$
On the other hand, $a\geq b\geq c$ so $(a-c)^2=(a-b)^2+(b-c)^2+2(a-b)(b-c)\geq (a-b)^2+(b-c)^2$
It suffices to show that $(S_a+S_b)(b-c)^2+(S_b+S_c)(a-b)^2\geq 0$
Because $S_b,S_c\geq 0$ so we only need to prove that $S_a+S_b\geq 0$
Or $\dfrac{1}{ab+bc+ca}\geq \dfrac{1}{(b+c)(c+a)}\Leftrightarrow c^2\geq 0$ which is obvious
The proof is completed, equalilty occur iff $a=b=c$

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mudok

#3 h Jun 9, 2016, 8:03 AM • 2 Y

Y by Adventure10, Mango247

Proof 2.
WLOG assume that $a\ge b\ge c$ , then it is easy to check $\frac{8abc}{(a+b)(b+c)(c+a)}\ge \frac{4ac}{(a+c)^2}$ . And

$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ac}{(a+c)^2}\ge 2 \ \ \iff \ \ \ (a^2+c^2-ab-bc)^2\ge 0$

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mudok

#5 h Jun 9, 2016, 8:09 AM • 2 Y

Y by Adventure10, Mango247

It is old inequality:

http://artofproblemsolving.com/community/c6h495781p2783722

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#6 h Jun 9, 2016, 11:25 AM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

Prove that $\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)}\ \ge\ 2\ .$

We have
$\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)} - 2 = \frac{\displaystyle \sum c(a+b-c)^2(a-b)^2}{(ab+bc+ca)(a+b)(b+c)(c+a)} \geqslant 0.$ The following inequality stronger is true
$\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)} \geqslant 2 + \frac{4(a-b)^2(b-c)^2(c-a)^2}{(a+b)^2(b+c)^2(c+a)^2},$ $\frac {a^2+b^2+c^2}{ab+bc+ca}+\frac {8abc}{(a+b)(b+c)(c+a)} \geqslant 2 + \frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)}.$

This post has been edited 2 times. Last edited by Nguyenhuyen_AG, Jun 9, 2016, 6:12 PM

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sqing

#7 h Jul 12, 2021, 10:47 PM

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For any $\{a,b,c\}\subset\mathrm R_+^*\ .$ Prove that $\frac{ab+bc+ca} {a^2+b^2+c^2}-\frac {4abc}{(a+b)(b+c)(c+a)}\ \le\frac{1}{2}.$ Where?
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This post has been edited 1 time. Last edited by sqing, Jul 13, 2021, 12:22 AM

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